Gutkin s Problem in Constant Curvature Geometries and Discrete Version

Transcription

1 Gutkin s Problem in Constant Curvature Geometries and Discrete Version Xidian Sun Yuwen Wang August 8th, 2013

2 Gutkin s Theorem in E 2 Given a smooth, convex and closed curve γ, assumethattwo points, X and Y, can chase each other around γ in such a way that the angle made by the chord XY with γ at both end points has a constant value, say, α. Ifγ is not a circle, what are possible values of α? The answer is as follows: a necessary and sucient condition is that Theorem There exists integer k 2 such that k tan α =tan(kα).

3 Motivation and Goal Motivation Billiards problem Rigidity theory The result can be also interpreted in terms of capillary floating with zero gravity in neutral equilibrium

4 Motivation and Goal Motivation Billiards problem Rigidity theory The result can be also interpreted in terms of capillary floating with zero gravity in neutral equilibrium Goal Develop similar behavior in discrete case Develop similar behavior in other constant curvature geometries

5 Gutkin s theorem in E 2 Figure : Curve γ with chord xy

6 What we know: γ is a convex, smooth, and closed curve. X = γ(x(t)) and Y = γ(y(t)) are two points on γ. The angles made by chord XY and γ are constant for all t. x(t), y(t) [0, L(γ)] are arc length parameters of γ.

7 What we know: γ is a convex, smooth, and closed curve. X = γ(x(t)) and Y = γ(y(t)) are two points on γ. The angles made by chord XY and γ are constant for all t. x(t), y(t) [0, L(γ)] are arc length parameters of γ. Want to find: How to parameterize x(t) andy(t)?

8 What we know: γ is a convex, smooth, and closed curve. X = γ(x(t)) and Y = γ(y(t)) are two points on γ. The angles made by chord XY and γ are constant for all t. x(t), y(t) [0, L(γ)] are arc length parameters of γ. Want to find: How to parameterize x(t) andy(t)? Define: L(x, y) be the arc length of chord XY.

9 Proposition We should choose t in a way such that x t = a/κ(x) and y t = a/κ(y), wherea is a constant.

10 Proof. We have following equations from Bialy s paper: L x = cos φ, L y = cos ψ L xx = sin2 φ κ(x)sinφ L yy = sin2 ψ κ(y)sinψ L L sin φ sin ψ L xy = L We want φ and ψ to be a constant α. Replacingα into above equations and since α is a constant, we have that: Then we can compute that 0=L xt = L xx x t + L xy y t 0=L yt = L xy x t + L yy y t y t = κ(x) x t κ(y).

11 Remark We know that 0 t T,whereT is the upper bound of t. We can choose a to make T to be 2π for later computation.

12 Proposition Set f (t + c) and f (t c) be such that f (t + c) = sin α κ(x) and f (t c) = sin α κ(y).thenwehavethat f (t + c)+f (t c) =a cot α (f (t + c) f (t c)).

13 Proof. From former equations, we can solve for L that Then we have that L = sin α κ(x) + sin α κ(y). L = f (t + c)+f (t c). It follows that L = f (t + c)+ f (t c), where f denotes derivative of f respect to t. Bychainrule,wehavethat Therefore, L = L x ẋ + L y ẏ = a cos α f (t + c) f (t c)). sin α f (t + c)+f (t c) =a cot α (f (t + c) f (t c)).

14 Fourier Expansion Since f (t) is a function with period of 2π, thus using Fourier expansion, we have that f (t) = b k e ikt,whereb k C, and b k = b k. It follows that f (t ± c) = b k e ±ikc e ikt f (t ± c) = b k ike ±ikc e ikt. Plugging into former proposition and equating both sides, we get: k tan α = a tan kc

15 Remark a =1in E 2 by Gauss-Bonnet Theorem. k 2 in E 2 by Fourier analysis. c = α in E 2

16 Spherical version of Gutkin s theorem In S 2, the computation is much more complicated for a general curve. We know that circle is always a solution to this problem. Therefore, we start from a circle γ 0 with radius R. Then,weare going to deform γ 0 and find infinitesimal solutions in a small neighborhood of the circle.

17 Spherical version of Gutkin s theorem Theorem Given a smooth, convex and closed curve γ which can be obtained by deforming a circle, assume that two points, X and Y,can chase each other around γ in such a way that the angle made by the chord XY with γ at both end points has a constant value, say, α. One sufficient and necessary condition is that 1 tan(kc) =k tan(c) a cos 2 c tan 2 R +1, where k is an integer with k 2, R is the radius of the circle we deform and cos α = cos c sin 2 R sin 2 c+cos 2 R.

18 New Equations Bialy s equations in S 2 : L xx = sin2 α tan L κ(x)sinα Bialy s equation in E 2 : L x = cos α, L y = cos α L xy = sin2 α sin L L x = cos φ, L y = cos ψ L yy = sin2 α tan L κ(y)sinα L xx = sin2 φ L κ(x)sinφ L yy = sin2 ψ L sin φ sin ψ L xy = L κ(y)sinψ

19 New Propostions In S 2 : Proposition We should choose parameter t in a way such that x t = a/ κ(x) 2 +sin 2 α,wherea is a constant to make T =2π.

20 New Propostions In S 2 : Proposition We should choose parameter t in a way such that x t = a/ κ(x) 2 +sin 2 α,wherea is a constant to make T =2π. In E 2 : Proposition We should choose t in a way such that x t = a/κ(x) and y t = a/κ(y), wherea is a constant.

21 New Propositions In S 2 : Proposition Let f (t + c) and f (t c) be such that tan f (t + c) = sin α κ(x) and tan f (t c) = sin α κ(y).then f (t + c)+f (t c) =a cot α sin f (t + c) sin f (t c)).

22 New Propositions In S 2 : Proposition Let f (t + c) and f (t c) be such that tan f (t + c) = sin α κ(x) and tan f (t c) = sin α κ(y).then f (t + c)+f (t c) =a cot α sin f (t + c) sin f (t c)). In E 2 : Proposition Set f (t + c) and f (t c) be such that f (t + c) = sin α κ(x) and f (t c) = sin α κ(y).thenwehavethat f (t + c)+f (t c) =a cot α (f (t + c) f (t c)).

23 Solving the equation in S 2 sin f is involved in the equation and therefore cannot solve it as in E 2.

24 Solving the equation in S 2 sin f is involved in the equation and therefore cannot solve it as in E 2. Circle is always a solution to this question.

25 Solving the equation in S 2 sin f is involved in the equation and therefore cannot solve it as in E 2. Circle is always a solution to this question. Start from a circle γ 0 with radius R and deform it to find infinitesimal solutions.

26 Solving the equation in S 2 sin f is involved in the equation and therefore cannot solve it as in E 2. Circle is always a solution to this question. Start from a circle γ 0 with radius R and deform it to find infinitesimal solutions. f 0 = cot 1 ( cot R sin α ), which is a constant.

27 Solving the equation in S 2 Then we deform the circle infinitesimally with f (t) =f 0 + g(t) and c = c + δ, keepingα fixed.it follows that g + + g = a cot α (sin(f 0 + g + ) sin(f 0 + g )). Applying Taylor expansion to sin(x + y) and ignoring terms which has power higher than 2, we have that sin(x + y) =sinx + y cos x. Then g + + g = a cot α cos f 0 (g + g ).

28 Solving the equation in S 2 Doing similar Fourier expansion as in the Euclidean case, we find that tan kc = k tan α a cos f 0 = where k is an integer. k tan α a cos[cot 1 ( cot R sin α )] = k a tan α tan 2 R sin 2 α +1,

29 Lemma In S 2, α and c satisfy the following equation: cos α = cos c sin 2 R sin 2 c + cos 2 R

30 Analogous formula in S 2 Applying the lemma to former equation, we have that 1 tan kc = k tan c a cos 2 c tan 2 R +1.

31 Remark Taking limit when R approaches 0, we have that tan(kc) = k a tan c. We know that T = 1 L(γ) a 0 κ(x) 2 +sin 2 α dx =2π. WhenR approaches 0, κ(x) approaches infinity, and therefore we have that T = 1 L(γ) a 0 κ(x) dx =2π. Ontheunitsphere,whenR approaches 0, it is relatively small to the sphere and locally conformal to E 2. Therefore, by Gauss-Bonnet theorem again we have that a = 1 and then we obtain the same formula tan kc = k tan c as in E 2.

32 (7,3) (8,4) (10,3) (10,4) (12,4) (12,5)